JEE Mains · Physics · STD 11 - 13. oscillations
Two particles \(A\) and \(B\) of equal masses are suspended from two massless springs of spring constants \(K _{1}\) and \(K _{2}\) respectively.If the maximum velocities during oscillations are equal, the ratio of the amplitude of \(A\) and \(B\) is
- A \(\frac{ K _{2}}{ K _{1}}\)
- B \(\frac{ K _{1}}{ K _{2}}\)
- C \(\sqrt{\frac{ K _{1}}{ K _{2}}}\)
- D \(\sqrt{\frac{ K _{2}}{ K _{1}}}\)
Answer & Solution
Correct Answer
(D) \(\sqrt{\frac{ K _{2}}{ K _{1}}}\)
Step-by-step Solution
Detailed explanation
\(A _{1} \omega_{1}= A _{2} \omega_{2}\) \(A_{1} \sqrt{\frac{k_{1}}{m}}=A_{2} \sqrt{\frac{k_{2}}{m}}\) \(\frac{ A _{1}}{ A _{2}}=\sqrt{\frac{ k _{2}}{ k _{1}}}\)
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